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The value of equilibrium constant of the...

The value of equilibrium constant of the reaction:
`HI_(g) Leftrightarrow 1/2 H_(2(g)) +1/2 I_(2(g))` is 8.0. The equilibrium constant of the reaction `H_(2(g))+I_(2(g)) Leftrightarrow 2HI_(g)` will be

A

`1//6`

B

`1//8`

C

`1//16`

D

`1//64`

Text Solution

Verified by Experts

The correct Answer is:
D
`HI_(g) Leftrightarrow 1//2H_(2(g)) +1//2I_(2(g))`
`i.e, K=([H_(2)]^(1//2) [I_(2)]^(1//2))/([HI])=8`
`H_(2(g)) +I_(2(g)) Leftrightarrow 2HI_(g)`
`K.=([HI]^2)/([H_(2)] [I_(2)])=(1/8)^2 rArr K.=1/(64)`
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